Suppose that P(x) is a polynomial of degree n such that P(k) = \frac {k}{k+1} for k = 0, 1, 2, …, n . Find the value of P(n+1). (Subjective Problem 176 from Test of Mathematics)

Discussion:

Consider an auxiliary polynomial g(x) = (x+1)P(x) – x . g(x) is an n+1 degree polynomial (as P(x) is n degree and we multiply (x+1) with it). We note that g(0) = g(1) = … = g(n) = 0  (as the given condition allows (k+1) P(k) – k = 0 for all k from 0 to n). Hence 0, 1, 2, … , n are the n+1 roots of g(x).

Therefore we may write g(x) = (x+1)P(x) – x = C(x)(x-1)(x-2)…(x-n) where C is a constant. Put x = -1. We get g(-1) = (-1+1)P(-1) – (-1) = C(-1)(-1-1)(-1-2)…(-1-n).

Thus 1 = C (-1)^{(n+1) } (n+1)! gives us the value of C. We put the value of C in the equation (x+1)P(x) – x = C(x)(x-1)(x-2)…(x-n) and replace x by n+1 to get the value of P(n+1).

(n+2)P(n+1) - (n+1) = \frac { (-1)^{(n+1)}}{(n+1)!} (n+1)(n)(n-1) ... (1) implying P(n+1) = \frac { (-1)^{(n+1)} + (n+1)}{(n+2)}